A) [M0 L2 T0]
B) [ML2T]
C) [ML0 T-1]
D) [M0 L2 T-1]
Correct Answer: A
Solution :
In the given equation, \[\frac{\alpha z}{k\theta }\]should be dimensionless. \[\therefore \] \[\alpha =\frac{k\theta }{z}\] \[\Rightarrow \] \[[\alpha ]=\frac{[M{{L}^{2}}{{T}^{-2}}{{K}^{-1}}]\times [K]}{[L]}=[ML{{T}^{-2}}]\]and \[p=\frac{\alpha }{\beta }\] \[\Rightarrow \]\[[\beta ]=\left[ \frac{\alpha }{p} \right]=\frac{[ML{{T}^{-2}}]}{[M{{L}^{-1}}{{T}^{-2}}]}=[{{M}^{0}}{{L}^{2}}{{T}^{0}}]\]
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